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Enthalpy, thermal energy, and kinetic vs. potential energy |
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| Sep14-05, 09:54 AM | #1 |
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Enthalpy, thermal energy, and kinetic vs. potential energy
Hi everyone,
None of my professors seem to be able to clearly explain what enthalpy is and its relation to energy. If possible, please confirm or correct my reasoning below: Energy is an abstract (yet quantifiable) term that describes the capacity to cause a change (do work) in a system (e.g., push something or cause atoms to stay close together in a molecule). Energy can exist in many forms including thermal (kinetic) and chemical (potential). Enthalpy (H) is NOT a form of energy, but instead it is a measure of the amount of energy stored (as potential energy????) in a system (e.g., energy in all the chemical bonds of a molecule). Since it is practically impossible to measure potential energy (all forms of potential energy, right????), we must look at released energy (kinetic energy?). Because of natural random and statistacly probable entropy, when no energy is applied to keep order in a system, it is released spontaneously as kinetic thermal energy, right? And this can be quantified by heat/temperature change. And this is what enthalpy (H) measures. Correct? Ok, some more thoughts. Enthalpy (H) is the measure of kinetic thermal energy, correct?? This is useful because it shows how much energy was originally stored as potential energy in a system, right? Sort of like working backwards to understand the original conditions. Ok, a few more ideas. First law of thermodynamics states that all energy can be interconverted (aka transduced). HOWEVER, thermal energy (and I suppose all types of kinetic energy, right?) are a unique case because they OFTEN (not always) are unable to be transduced back into other forms of "useful to do work" energy. This is because kinetic energy naturally dissipates (spreads out) and, since the universe is still expanding, it dissipates out and is often "lost" because it fails to concentrate. In other cases, kinetic thermal energy can be used to power stuff (like a piston in a car, I think). Super sorry for the length, but I am very curious about this stuff. Please tell me if my reasoning makes sense and/or where I went wrong. Thanks in advance. |
| Sep14-05, 07:30 PM | #2 |
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Not a single response???
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| Sep14-05, 08:05 PM | #3 |
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[tex]H\equiv U + PV[/tex] because no other work modes are present, dU can be expressed as: [tex]dU=dQ + dw=dQ -pdV[/tex] Now, the change in enthalpy is: [tex]dH = dU + PdV +VdP= dQ -PdV +PdV -VdP[/tex] The process takes place at constant P, so: [tex]dH=dQ[/tex] So this is why the enthalpy is often considered: the change in enthalpy is the heat of the reaction, when carried out at constant P. |
| Sep14-05, 11:20 PM | #4 |
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Enthalpy, thermal energy, and kinetic vs. potential energy
Supplementing what LeonhardEuler wrote, see a discussion of the first law of thermodynamics and related concepts at http://hyperphysics.phy-astr.gsu.edu...firlaw.html#c1
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| Nov19-10, 07:14 AM | #5 |
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it is much easier to consider enthalpy as a form of energy, as explaned in the attachment.
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| Nov19-10, 08:18 AM | #6 |
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unfortunately the attachment doesn't show. I try it again.
I have a problem with a 'security token', whatever that may be. |
| Nov24-10, 06:39 AM | #7 |
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Well, 3rd try. Looks good this time: this attachment is about enthalpy.
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